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(a) These are the masses, in kg, of ten items.
38 6 21 50 32 7 15 9 27 25
The maximum total mass that each bin can hold is 60 kg. Amara uses Method 1 to put these ten items into bins. The table shows how she puts the first 6 items into bins.
[Table below]
Bin | Mass of items in bin | Unused mass in bin
1 | 38, 6, 7 | 22, 16, 9
2 | 21, 32 | 39, 7
3 | 50 | 10
4 |
5 |
(i) Complete Amara’s table to show that she needs 5 bins. [4]
(ii) Work out the total unused mass in the 5 bins. [2] ..................................................
(b) These are the masses, in kg, of six items.
8 16 13 10 5 3
The maximum total mass that each bin can hold is 20 kg.
[Table below]
Bin | Mass of items in bin | Unused mass in bin
1 | 8 | 12
2 |
3 |
4 |
5 |
Use Method 1 to complete the table for all six items. The first item has been put in for you. You may not need all the bins. [2]
538
551
581
Amara wants to see if she can use fewer bins. She puts her items in order of mass before she puts them in bins. She uses this method.
Method 2 Put the masses in order, largest first. Then use Method 1.
These are the masses, in kg, of the ten items from Question 1(a).
38 6 21 50 32 7 15 9 27 25
(a) Write these ten masses in order, largest first.
......... , ......... , ......... , ......... , ......... , ......... , ......... , ......... , ......... , ......... [1]
(b) The maximum total mass that each bin can hold is 60 kg. Complete the table using Method 2.
[Table_1]
................................................................................................................. [3]
(c) Work out the difference in the total unused mass when using Method 1 and Method 2. Use your answers from Question 1(a)(ii) and Question 2(b).
.................................................. [2]
573
599
553
A \textit{best solution} uses the smallest possible number of bins.
(a) (i) A set of items with a total mass of 270 kg is put into 4 bins. The maximum total mass that each bin can hold is 80 kg.
Show that this is a best solution. [2]
(ii) Show that the solution in Question 1(b) is a best solution. [2]
(b) Amara knows that for a particular set of items a best solution is 6 bins. The maximum total mass that each bin can hold is 5 kg. The total mass of the items is 27.5 kg.
Work out the amount of unused storage for a best solution for these items. [2]
514
535
543
Amara tries another way to improve Method 1.
Method 3 Look for items that combine to make as many full bins as possible and place these first. For the remaining items, use Method 2.
(a) These are the masses, in kg, of eight items.
21 10 30 19 13 7 28 4
The maximum total mass that each bin can hold is 40 kg.
Does Method 3 give a best solution for these items? Show how you decide.
| Bin | Mass of items in bin | Unused mass in bin |
| --- | --------------------- | ------------------- |
| 1 | | |
| 2 | | |
| 3 | | |
| 4 | | |
| 5 | | |
(b) Amara puts nine items into bins using Method 3. The maximum total mass that each bin can hold is 40 kg.
| Bin | Mass of items in bin | Unused mass in bin |
| --- | --------------------- | ------------------- |
| 1 | 18, 22 | 0 |
| 2 | 32, 5, 3 | 0 |
| 3 | 32 | 8 |
| 4 | 19, 15 | 6 |
| 5 | 12 | 28 |
Amara only wants to use 4 bins. She removes the last item she packed and divides it into two smaller items with the same total mass. She puts each of these two items into a bin that can hold its mass.
Work out how much the percentage of unused storage changes when Amara uses 4 bins instead of 5 bins.
541
559
507
These are the masses, in kg, of eight items.
31 10 39 20 29 47 50 12
The maximum total mass that each bin can hold is $60 ext{ kg}$.
Each bin Amara uses costs $13.50.
Use Method 2 or Method 3 to put these items into bins to give a best solution.
Find the cost of this solution.
[Table_1]
| Bin | Mass of items in bin | Unused mass in bin |
|-----|----------------------|-------------------|
| 1 | | |
| 2 | | |
| 3 | | |
| 4 | | |
| 5 | | |
$ ext{..........................................}
523
520
523
